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Q.Find the mean deviation about median for the following frequency distribution. | Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 5 | 8 | 15 | 16 | 6 |

Assam AhsecAHSEC Higher Secondary (HS) 1st Year Examination 2026Subjective· 5mImportance★★★★★
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Median class 2020–3030 gives median =28=28; M.D.=∑fi∣xi−28∣N=47850=9.56\text{M.D.}=\frac{\sum f_i|x_i-28|}{N}=\frac{478}{50}=9.56.

Data (N=5+8+15+16+6=50N = 5+8+15+16+6 = 50):

| Class | ff | Mid xx | c.f. |

| 0-10 | 5 | 5 | 5 |

| 10-20 | 8 | 15 | 13 |

| 20-30 | 15 | 25 | 28 |

| 30-40 | 16 | 35 | 44 |

| 40-50 | 6 | 45 | 50 |

Median: N2=25\frac{N}{2} = 25. The c.f. first reaches/exceeds 2525 in class 2020–3030, so median class is 2020–3030: l=20, c.f. before=13, f=15, h=10l=20,\ \text{c.f. before}=13,\ f=15,\ h=10.

Median=l+N2−c.f.f×h=20+25−1315×10=20+12015=20+8=28\text{Median} = l + \dfrac{\frac{N}{2} - \text{c.f.}}{f}\times h = 20 + \dfrac{25-13}{15}\times10 = 20 + \dfrac{120}{15} = 20 + 8 = 28.

Mean deviation about median: compute fi∣xi−28∣f_i|x_i - 28|:

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